Question

For $n\ge 2$, let $C_r=\left(\begin{array}{c}n\\ r\end{array}\right)$ and $a_n={\sum }_{r=0}^n\frac{1}{C_r}$

Answer 1

A) Using $C_r=C_{n-r}$
${\sum }_{r=0}^n\frac{r}{C_r}={\sum }_{r=0}^n\frac{r}{C_{n-r}}={\sum }_{r=0}^n\frac{n-r}{C_r}={\sum }_{r=0}^n\frac{n}{C_r}-{\sum }_{r=0}^n\frac{r}{C_r}\Rightarrow 2{\sum }_{r=0}^n\frac{r}{C_{n-r}}={\sum }_{r=0}^n\frac{n}{C_r}=n{a}_n\Rightarrow {\sum }_{r=0}^n\frac{r}{C_{n-r}}={\sum }_{r=0}^n\frac{r}{C_r}=\frac{1}{2}n{a}_n$
B) ${\sum }_{r=0}^n\frac{n-r}{C_{n-r}}={\sum }_{r=0}^n\frac{n}{C_r}-{\sum }_{r=0}^n\frac{r}{C_r}=n{a}_n-\frac{1}{2}n{a}_n=\frac{1}{2}n{a}_n$
C) Using ${rC}_r=n{C}_{r-1}$
${\sum }_{r=0}^n\frac{1}{{rC}_r}=\frac{1}{n}{\sum }_{r=0}^n\frac{1}{C_{r-1}}=\frac{1}{n}{\sum }_{r=0}^n\frac{1}{C_r}=\frac{1}{n}{a}_{n-1}$
D) ${\sum }_{r=0}^{n-1}\frac{1}{{(n-r)C}_r}={\sum }_{r=0}^{n-1}\frac{1}{{n{C}_r-rC}_r}={\sum }_{r=0}^{n-1}\frac{1}{n{C}_r-n{C}_{r-1}}=\frac{1}{n}{a}_{n-1}$